Multivariable Advanced Calculus

278 BROUWER DEGREEProof: Let B (y,3δ) ∩ g (f (∂Ω)) = ∅ and let ˜g ∈ C 2 (R n , R n ) be such thatsup { |˜g (z) − g (z)| : z ∈ f ( Ω )} < δAnd also y is a regular value of ˜g. Then from the above inequality, if x ∈ ∂Ω andt ∈ [0, 1],|g (f (x)) + t (˜g (f (x)) − g (f (x))) − y| ≥ |g (f (x)) − y| − t |˜g (f (x)) − g (f (x))|≥ 3δ − tδ > 0.It follows thatd (g ◦ f, Ω, y) = d (˜g ◦ f, Ω, y) . (10.17)Now also, ∂K i ⊆ f (∂Ω) and so if z ∈ ∂K i , then g (z) ∈ g (f (∂Ω)). Consequently, forsuch z,|g (z) + t (˜g (z) − g (z)) − y| ≥ |g (z) − y| − tδ > 3δ − tδ > 0which shows that, by homotopy invariance,Therefore, by Lemma 10.5.3,d (g ◦ f, Ω, y) = d (˜g ◦ f, Ω, y) ==d (g, K i , y) = d (˜g, K i , y) . (10.18)∞∑d (˜g, K i , y) d (f, Ω, K i )i=1∞∑d (g, K i , y) d (f, Ω, K i )i=1and the sum has only finitely many non zero terms. This proves the product formula.Note there are no convergence problems because these sums are actually finite sumsbecause, as in the previous lemma, g −1 (y) ∩ f ( Ω ) is a compact set covered by thecomponents of R n \ f (∂Ω) and so it is covered by finitely many of these components.For the other components, d (f, Ω, K i ) = 0 or else d (g, K i , y) = 0. The following theorem is the Jordan separation theorem, a major result. A homeomorphismis a function which is one to one onto and continuous having continuousinverse. Before the theorem, here is a helpful lemma.Lemma 10.5.5 Let Ω be a bounded open set in R n , f ∈ C ( Ω; R n) , and suppose{Ω i } ∞ i=1are disjoint open sets contained in Ω such thaty /∈ f ( Ω \ ∪ ∞ )j=1Ω jThend (f, Ω, y) =∞∑d (f, Ω j , y)j=1where the sum has all but finitely many terms equal to 0.Proof: By assumption, the compact set f −1 (y) has empty intersection withΩ \ ∪ ∞ j=1Ω jand so this compact set is covered by finitely many of the Ω j , say {Ω 1 , · · · , Ω n−1 } andy /∈ f ( ∪ ∞ j=nΩ j)